Sodium metal crystallizes in a body centred cubic lattice with unit cell edge length of 4 Aᵒ. The radius of sodium atom is __________ × 10⁻¹ Aᵒ (Nearest integer)

Solution

Chemistry ma’am · 4 steps

In bcc, the atoms touch along the body diagonal, so set 4r equal to √3 a.

1. Where atoms touch. In a body centred cubic cell, the corner atoms do not touch each other along the edge. They touch the centre atom along the body diagonal, which runs from one corner through the centre to the opposite corner.
2. Count the radii. That diagonal covers one corner atom's radius, the full centre atom (2r) and the opposite corner atom's radius, so it equals 4r. Its length is √3 a.

   4r = √3 a ⇒ r = √3/4 a

3. Put in numbers. With a = 4 Å, r = √3/4 × 4 = √3 = 1.732 Å.
4. So. r = 17.32 × 10⁻¹ Å, which rounds to 17.

The common slip: 14. Using the fcc relation 4r = √2 a gives r = 1.414 Å, that is 14, not 17. In bcc the atoms touch along the body diagonal, not the face diagonal.

The idea to keep. First find the line along which atoms actually touch: the edge in simple cubic (a = 2r), the face diagonal in fcc (√2a = 4r), the body diagonal in bcc (√3a = 4r).

Answer from the NTA final answer key. Solution by PrepWiser.